Given: klmn is a trapezoid, kl=mn, m∠1=m∠2, lm/kn = 8/9 , perimeter klmn=132 find: the length of midsegment.

34

KM is a transversal relative to parallel lines LM and KN. Thus ∠2 = ∠MKN ≅ ∠KML and ∠KML = ∠1. The two base angles of ΔKLM are equal, so that triangle is isosceles.

Then the ratios of all the sides are ...

... KL : LM : MN : KN = 8 : 8 : 8 : 9

The sum of these ratio units is 33, so each one stands for 132/33 = 4 perimeter length units. Then segment LM is 8×4 = 32 perimeter length units, and KN is 9×4 = 36 permeter length units.

The midsegment is the average of lengths LM and KN, so is ...

... (32 +36)/2 = 34 . . . . perimeter length units

The length of midsegment is 34 units.

Given data:

The trapezoid KLMN, Such that KL = MN.

And , LM/KN = 8/9

Also, perimeter of KLMN = 132 units.

To find:

The length of midsegment (KM).

In the given problem, we can observe that KM is a transversal relative to parallel lines LM and KN. Which means,

Clearly, two base angles are equal. So, the triangles KLM and KMN are isosceles.

Taking the ratios of sides of two triangles as,

= KL : LM : MN : KN

= 8 : 8 : 8 : 9

The sum of ratio units is, 8 + 8 +8 +9 = 33. Then, the value of each ratio is,

Then the length of segment LM is,

And, length of segment KN is,

Then, the length of midsegment KM is obtained by taking the average of LM and KN as,

Thus, the length of midsegment is 34 units.

Learn more about the concept of midsegments here:

link

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.

150.6

all i did was i multiplied 5.02 by 30